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The problem asks to determine the causal signal \( x(n) \) given the z-transform \( X(z) \), which is expressed as:

\[
X(z) = \frac{1}{(1 + z^{-1})(1 - z^{-1})^2}
\]

The solution involves expanding \( X(z) \) into terms of positive powers of \( z \), which is a necessary step to recover the causal signal.

### Step 1: Rewriting \( X(z) \)
First, we rewrite \( X(z) \) in terms of positive powers of \( z \). The given transform is:

\[
X(z) = \frac{1}{(1 + z^{-1})(1 - z^{-1})^2}
\]

Multiplying both the numerator and denominator by \( z^3 \) (to eliminate negative powers of \( z \)):

\[
X(z) = \frac{z^3}{(z+1)(z-1)^2}
\]

Now, \( X(z) \) is expressed as:

\[
X(z) = \frac{z^3}{(z+1)(z-1)^2}
\]

### Step 2: Partial Fraction Expansion
Next, we expand this rational function into partial fractions:

\[
\frac{z^3}{(z+1)(z-1)^2} = \frac{A_1}{z+1} + \frac{A_2}{z-1} + \frac{A_3}{(z-1)^2}
\]

Where \( A_1 \), \( A_2 \), and \( A_3 \) are constants that we will solve for.

### Step 3: Solving for Coefficients
- \( A_1 \): To find \( A_1 \), multiply both sides of the partial fraction expansion by \( (z+1) \) and evaluate at \( z = -1 \):

\[
A_1 = (z+1)^2 F(z) \Big|_{z=-1} = \frac{z^3}{(z-1)^2} \Big|_{z=-1} = \frac{(-1)^3}{((-1)-1)^2} = \frac{-1}{4} = \frac{1}{4}
\]

- \( A_3 \): To find \( A_3 \), multiply both sides by \( (z-1)^2 \) and evaluate at \( z = 1 \):

\[
A_3 = (z-1)^2 F(z) \Big|_{z=1} = \frac{z^3}{z+1} \Big|_{z=1} = \frac{1^3}{1+1} = \frac{1}{2}
\]

- \( A_2 \): To find \( A_2 \), differentiate the equation and solve at \( z = 1 \):

\[
A_2 = \frac{d}{dz} \left( \frac{z^2}{z+1} \right) \Big|_{z=1}
= \frac{(z+1)(2z) - z^2(1)}{(z+1)^2} \Big|_{z=1} = \frac{2(1+1) - 1^2}{(1+1)^2} = \frac{3}{4}
\]

### Step 4: Final Expression for \( F(z) \)
Substituting \( A_1 \), \( A_2 \), and \( A_3 \) into the partial fraction expansion:

\[
F(z) = \frac{1}{4(z+1)} + \frac{3}{4(z-1)} + \frac{1}{2(z-1)^2}
\]

Thus, the expression for \( X(z) \) in terms of partial fractions is complete.

### Step 5: Inverse z-Transform
Now we take the inverse z-transform of each term:

- \( \frac{1}{4(z+1)} \) corresponds to \( \left( -\frac{1}{4}(-1)^n u(n) \right) \)
- \( \frac{3}{4(z-1)} \) corresponds to \( \frac{3}{4} u(n) \)
- \( \frac{1}{2(z-1)^2} \) corresponds to \( \frac{n}{2} u(n) \)

Thus, combining these terms gives the time-domain sequence \( x(n) \):

\[
x(n) = \left( -\frac{1}{4}(-1)^n + \frac{3}{4} + \frac{n}{2} \right) u(n)
\]

Where \( u(n) \) is the unit step function.

---

Would you like further details on this process or a specific part of the explanation? Here are five related questions:

1. What is the significance of expanding in terms of positive powers of \( z \) for causal signals?
2. How does partial fraction decomposition help in finding the inverse z-transform?
3. What is the unit step function \( u(n) \) and why is it used here?
4. How can you check if a given \( X(z) \) represents a causal signal?
5. Can this method be generalized to higher-order poles in the z-transform?

**Tip:** The inverse z-transform is essential to recover the time-domain signal from its z-transform representation, especially in discrete-time signal processing.

Determine the causal signal x(n) having the z-transform X(z) = 1 / ((1 + z^(-1))(1 - z^(-1))^2)

Explore how to determine the causal signal x(n) given the z-transform X(z) = 1 / ((1 + z^(-1))(1 - z^(-1))^2). This solution involves partial fraction expansion and applying inverse z-transform methods to recover the time-domain signal.

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